3D Reading

• what does invertible mean w.r.t. linear maps?

title: invertible, inverse
- A linear map $T \in L(V,W)$ is called *invertible* if there exists a linear map $S \in L(W,V)$ such that $ST$ equals the identity map on $V$ and $TS$ equals the identity map on $W$. 
- A linear map $S \in L(W,V)$ satisfying $ST = I$ and $TS=I$ is called an *inverse* of $T$ (note that the first $I$ is the identity map on $V$ and the second $I$ is the identity map on $W$.

relates to injectivity and surjectivity.. #extract

title: inverse is unique
An invertible linear map has a unique inverse

proof is just setting \[S_1 = S_1I = S_1(TS_2) ... = S_2\] long chain of equalitys based on the fact it acts as the inverse

title: $T^{-1}$
If $T$ is invertible, then its inverse is denoted by $T^{-1}$. In other words, if $T \in L(V,W)$ is invertible, then $T^{-1}$ is the unique element of $L(W,V)$ such that $T^{-1}T=I$ and $TT^{-1}=I$.

Just the inverse. oh, and here we get the thing i was wondering about earlier

title: invertibility is equivalent to injectivty and surjectivity
A linear map is invertivle iff it is injective and surjective

aka. bijective means invertible

pretty simple proof, just using the definitions very closely.

• multi by \(x^2\) from P(R) to P(R) cus not surjective. #extract
  • this is because 1 is not in the range.
• backwards shift from f^infin to itself
  • not injective

'essentially the same,' bar the names of the elems

title: isomorphism, isomorphic
- An isomorphism is an invertible linear map
- two vector spaces are called isomorphic if there is an isomorphism from one vector space onto the other one

which means they are bijective.

essentially relabeling \(v\) as \(Tv\).

isomorphic means equal shape in greek
isos -> equal, morph -> shape

but people use isomprhism when they want to emphazise how two spaces can be essentially the same

title: dimension shows whether vector spaces are isomorphic
Two finite-dim vec spaces over **F** are isomorphic iff they have the same dimension

makes sense.

uh,

title: L(V,W) and $F^(m,n)$ are isomorphic
Suppose $v_1, \dots, v_n$ is a basis of $V$ and $w_1, \dots, w_m$ is a basis of $W$. Then $M$ is an isomprhism between $L(V,W)$ and $F^{m,n}$.

amazing.

pluging in dimension results, we get

title: $\dim L(V,W)=(\dim V)(\dim W)$
Suppose $V$ and $W$ are finite-dim. Then $L(V,W)$ is finite-dim and 
$$\dim L(V,W)=(\dim V)(\dim W)$$

Amazing proof.

not the matrix of a map like in KBxChapter3CReading, but the matrix of a vector.

title: matrix of a vector, M(v)
Suppose $v \in V$ and $v_1, \dots, v_n$ is a basis of $V$. The matrix of v w.r.t. this basis is the n-by-1 matrix 
$$
M(v) = \begin{bmatrix} 
c_1  \\ 
 \vdots \\ 
 c_n  
 \end{bmatrix}.
$$

where $c_1, \dots, c_n$ are the scalars such that
$$ v = c_1 v_1 + \dots + c_n V_n. $$

again, about representing as a KBxLinearCombinations of the basis vecs.

title: $M(T)._{,k} = M(vk).$
Suppose $T \in L(V,W)$ and $v_1, \dots, v_n$ is a basis of $V$ and $w_1, \dots, w_m$ is a basis of $W$. Let 1 <= k <= n. Then the $k^{th}$ column of $M(T)$, which is denoted by $M(T)._{,k}$ equals $M(vk)$.

takin their word for this one.

now we fit together mat of lin map, mat of vec, and matmul.

title: linear maps act like matrix multiplication
Suppose $T \in L(V,W)$ and $v \in V$. Suppose $v_1, \dots, v_n$ is a basis of $V$ and $w_1, \dots, w_m$ is a basis of $W$. Then 
$$M(Tv) = M(T)M(v).$$

every m-by-n mat induces another linear map from \(F^{n,1}\) to \(F^{1,m}.\)

with isomorphisms, we can think of linear maps as multiplicaions on \(F^{n,1}\) by some mat.

axler focuses on linear maps as matricie and vise versa

linear maps from a vector space to itself are very important

we give them their own notation,

title: operator, $L(V)$
- A linear map from a vector space to itself is called an *operator.*
- The notation $L(V)$ denotes the set of all operators on $V$. In other words, $L(V) = L(V,V).$

THIS IS IMPORTANT! says that the deepest and more important parts of linalg (and the rest of the book) deal with operators. #extract

• remember, neither injectivity nor surjectivity implies invertibility
• but it does for operators! normally, injectivity is easier so that is checked

title: injectivity is equivalent to surjectivity in finite dimensions
Suppose $V$ is finite-dim and $T \in L(V).$ Then the following are equivalent:

a. T is invertible
b. T is injective
c. T is surjective

proof via multiple uses of fundemental theorem of linear algebra

this seems like a very important chapter. need to do some extracting on this one.
big concepts were isomorphism and operators. extract later!